another identity

**windir** · August 9th 2009, 03:53 PM

Prove...
$\frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

this is as far as i got
working from LHS

$\frac{1}{sec(x)-tan(x)}$

$cos(x)-cot(x)$

$cos(x)-\frac{cos(x)}{sin(x)}$

$\frac{sin(x)}{sin(x)}cos(x)-\frac{cos(x)}{sin(x)}$

$\frac{sin(x)cos(x)}{sin(x)}-\frac{cos(x)}{sin(x)}$

now i'm stuck..

**Bruno J.** · August 9th 2009, 04:05 PM

You must know that $(a-b)(a+b)=a^2-b^2$ . This identity is worth being very familiar with.

So you want to prove $1=\sec^2x-\tan^2x$ (I just multiplied both sides of your identity by $\sec x - \tan x$ ).
Now just apply the identity $\sec^2x=1+\tan^2x$ and you're done!

**Chris L T521** · August 9th 2009, 04:06 PM

Originally Posted by windir

Prove...
$\frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

this is as far as i got
working from LHS

$\frac{1}{sec(x)-tan(x)}$

$cos(x)-cot(x)$

$cos(x)-\frac{cos(x)}{sin(x)}$

$\frac{sin(x)}{sin(x)}cos(x)-\frac{cos(x)}{sin(x)}$

$\frac{sin(x)cos(x)}{sin(x)}-\frac{cos(x)}{sin(x)}$

now i'm stuck..

$\frac{1}{\sec x-\tan x}{\color{red}\neq}\frac{1}{\sec x}-\frac{1}{\tan x}$ !!!! XD

What I suggest you do with the LHS is the following: $\frac{1}{\sec x-\tan x}=\frac{1}{\sec x-\tan x}\cdot\frac{\sec x+\tan x}{\sec x+\tan x}=\dots$

In the process, invoke one of the pythagorean identities...

Can you take it from here?

**Random Variable** · August 9th 2009, 04:08 PM

$\frac{1}{\sec x - \tan x} \cdot \frac{\sec x + \tan x}{\sec x + \tan x}$

$= \frac {\sec x + \tan x}{\sec^{2} x - \tan^{2} x }$

$= \frac {\sec x + \tan x}{1 + \tan^{2}x - \tan^{2}x}$

$= \sec x + \tan x$

**windir** · August 9th 2009, 04:13 PM

ooooooooooohhh, duh! thanks man!

**mr fantastic** · August 11th 2009, 04:11 AM

Originally Posted by windir

Prove...
$\frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

[snip]

Also discussed here: http://www.mathhelpforum.com/math-he...589-proof.html

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